Question: An ellipse is drawn with major and minor axes of lengths 10 and 8 respectively.  Using one focus as a center, a circle is drawn that is tangent to the ellipse, with no part of the circle being outside the ellipse.  Compute the radius of the circle.
Explanation: Place the ellipse in the coordinate plane, as usual, so that the center is at the origin.  Then the equation of the ellipse is
\[\frac{x^2}{25} + \frac{y^2}{16} = 1.\]Also, the distance from the center to each foci is $\sqrt{5^2 - 4^2} = 3,$ so one foci is at $F = (3,0).$

[asy]
unitsize(0.6 cm);

path ell = xscale(5)*yscale(4)*Circle((0,0),1);
pair F = (3,0);

draw(ell);
draw(Circle(F,2));
draw((-5,0)--(5,0));
draw((0,-4)--(0,4));

dot("$F = (3,0)$", F, S);
[/asy]

Consider the circle centered at $F$ with radius 2.  The equation of this circle is $(x - 3)^2 + y^2 = 4,$ so $y^2 = 4 - (x - 3)^2.$  Substituting into the equation of the ellipse, we get
\[\frac{x^2}{25} + \frac{4 - (x - 3)^2}{16} = 1.\]This simplifies to $3x^2 - 50x + 175 = 0,$ which factors as $(x - 5)(3x - 35) = 0.$  The solutions are $x = 5$ and $x = \frac{35}{3},$ the latter root being extraneous.  This tells us that the ellipse and circle intersect only at the point $(5,0),$ and clearly we cannot draw a larger circle.

Hence, the maximum radius is $\boxed{2}.$